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Quantas instituições existem? | CREATE TABLE inst (Id VARCHAR) | SELECT COUNT(*) FROM inst |
Quantos artigos são publicados no total? | CREATE TABLE papers (Id VARCHAR) | SELECT COUNT(*) FROM papers |
Quais são os títulos dos artigos publicados por "Jeremy Gibbons"? | CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR) | SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Jeremy" AND t1.lname = "Gibbons" |
Encontre todos os artigos publicados pelo "Aaron Turon". | CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR) | SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Aaron" AND t1.lname = "Turon" |
Quantos artigos publicou o "Atsushi Ohori"? | CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR); CREATE TABLE papers (paperid VARCHAR) | SELECT COUNT(*) FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Atsushi" AND t1.lname = "Ohori" |
Qual é o nome da instituição a que pertence "Matthias Blume"? | CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR); CREATE TABLE inst (name VARCHAR, instid VARCHAR) | SELECT DISTINCT t3.name FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t1.fname = "Matthias" AND t1.lname = "Blume" |
A que instituição pertence "Katsuhiro Ueno"? | CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR); CREATE TABLE inst (name VARCHAR, instid VARCHAR) | SELECT DISTINCT t3.name FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t1.fname = "Katsuhiro" AND t1.lname = "Ueno" |
Quem pertence à instituição "Universidade de Oxford"? | CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (fname VARCHAR, lname VARCHAR, authid VARCHAR); CREATE TABLE inst (instid VARCHAR, name VARCHAR) | SELECT DISTINCT t1.fname, t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "University of Oxford" |
Quais autores pertencem à instituição "Google"? | CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (fname VARCHAR, lname VARCHAR, authid VARCHAR); CREATE TABLE inst (instid VARCHAR, name VARCHAR) | SELECT DISTINCT t1.fname, t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Google" |
Quais são os sobrenomes do autor do artigo intitulado "Binders Unbound"? | CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (lname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR) | SELECT t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title = "Binders Unbound" |
Encontre o nome e o apelido do autor (s) que escreveu o artigo "Não tem nome, não sente dor". | CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR); CREATE TABLE authors (fname VARCHAR, lname VARCHAR, authid VARCHAR) | SELECT t1.fname, t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title = "Nameless , Painless" |
Quais são os trabalhos publicados na instituição "Universidade de Indiana"? | CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR) | SELECT DISTINCT t1.title FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Indiana University" |
Encontre todos os artigos publicados pela instituição "Google". | CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR) | SELECT DISTINCT t1.title FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Google" |
Quantos trabalhos são publicados pela instituição "Universidade de Tokohu"? | CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR) | SELECT COUNT(DISTINCT t1.title) FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Tokohu University" |
Encontre o número de artigos publicados pela instituição "Universidade da Pensilvânia". | CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR) | SELECT COUNT(DISTINCT t1.title) FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "University of Pennsylvania" |
Encontre os papéis que tenham "Olin Shivers" como autor. | CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR) | SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Olin" AND t1.lname = "Shivers" |
Que jornais têm "Stephanie Weirich" como autora? | CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR) | SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Stephanie" AND t1.lname = "Weirich" |
Que artigo é publicado em uma instituição nos "EUA" e tem "Turon" como seu segundo autor? | CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR, instid VARCHAR, authorder VARCHAR); CREATE TABLE authors (authid VARCHAR, lname VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE inst (instid VARCHAR, country VARCHAR) | SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid JOIN inst AS t4 ON t2.instid = t4.instid WHERE t4.country = "USA" AND t2.authorder = 2 AND t1.lname = "Turon" |
Encontre os títulos de artigos cujo primeiro autor é afiliado a uma instituição no país "Japão" e tem o sobrenome "Ohori"? | CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR, instid VARCHAR, authorder VARCHAR); CREATE TABLE authors (authid VARCHAR, lname VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE inst (instid VARCHAR, country VARCHAR) | SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid JOIN inst AS t4 ON t2.instid = t4.instid WHERE t4.country = "Japan" AND t2.authorder = 1 AND t1.lname = "Ohori" |
Qual é o nome do autor que mais trabalhos publicou? | CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (lname VARCHAR, fname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR) | SELECT t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.fname, t1.lname ORDER BY COUNT(*) DESC LIMIT 1 |
Retire o país que publicou mais jornais. | CREATE TABLE inst (country VARCHAR, instid VARCHAR); CREATE TABLE authorship (instid VARCHAR, paperid VARCHAR); CREATE TABLE papers (paperid VARCHAR) | SELECT t1.country FROM inst AS t1 JOIN authorship AS t2 ON t1.instid = t2.instid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.country ORDER BY COUNT(*) DESC LIMIT 1 |
Encontre o nome da organização que publicou o maior número de artigos. | CREATE TABLE inst (name VARCHAR, instid VARCHAR); CREATE TABLE authorship (instid VARCHAR, paperid VARCHAR); CREATE TABLE papers (paperid VARCHAR) | SELECT t1.name FROM inst AS t1 JOIN authorship AS t2 ON t1.instid = t2.instid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.name ORDER BY COUNT(*) DESC LIMIT 1 |
Encontre os títulos dos artigos que contêm a palavra "ML". | CREATE TABLE papers (title VARCHAR) | SELECT title FROM papers WHERE title LIKE "%ML%" |
Qual artigo tem o título "Database"? | CREATE TABLE papers (title VARCHAR) | SELECT title FROM papers WHERE title LIKE "%Database%" |
Encontre os nomes de todos os autores que escreveram um artigo com o título contendo a palavra "Funcional". | CREATE TABLE authors (fname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR); CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR) | SELECT t1.fname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title LIKE "%Functional%" |
Encontre os sobrenomes de todos os autores que escreveram um artigo com título contendo a palavra "monádico". | CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (lname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR) | SELECT t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title LIKE "%Monadic%" |
Retire o título do artigo que tem o maior número de autores. | CREATE TABLE authorship (authorder INTEGER); CREATE TABLE authorship (paperid VARCHAR, authorder INTEGER); CREATE TABLE papers (title VARCHAR, paperid VARCHAR) | SELECT t2.title FROM authorship AS t1 JOIN papers AS t2 ON t1.paperid = t2.paperid WHERE t1.authorder = (SELECT MAX(authorder) FROM authorship) |
Qual é o primeiro nome do autor com o sobrenome "Ueno"? | CREATE TABLE authors (fname VARCHAR, lname VARCHAR) | SELECT fname FROM authors WHERE lname = "Ueno" |
Encontre o sobrenome do autor com o nome "Amal". | CREATE TABLE authors (lname VARCHAR, fname VARCHAR) | SELECT lname FROM authors WHERE fname = "Amal" |
Encontre os nomes dos autores, ordenados por ordem alfabética. | CREATE TABLE authors (fname VARCHAR) | SELECT fname FROM authors ORDER BY fname |
Retire todos os sobrenomes dos autores em ordem alfabética. | CREATE TABLE authors (lname VARCHAR) | SELECT lname FROM authors ORDER BY lname |
Retirada todos os nomes e apelidos dos autores na ordem alfabética dos apelidos. | CREATE TABLE authors (fname VARCHAR, lname VARCHAR) | SELECT fname, lname FROM authors ORDER BY lname |
Quantos nomes diferentes têm os atores e as atrizes? | CREATE TABLE actor (last_name VARCHAR) | SELECT COUNT(DISTINCT last_name) FROM actor |
Qual é o primeiro nome mais popular dos atores? | CREATE TABLE actor (first_name VARCHAR) | SELECT first_name FROM actor GROUP BY first_name ORDER BY COUNT(*) DESC LIMIT 1 |
Qual é o nome completo mais popular dos atores? | CREATE TABLE actor (first_name VARCHAR, last_name VARCHAR) | SELECT first_name, last_name FROM actor GROUP BY first_name, last_name ORDER BY COUNT(*) DESC LIMIT 1 |
Que distritos têm pelo menos dois endereços? | CREATE TABLE address (district VARCHAR) | SELECT district FROM address GROUP BY district HAVING COUNT(*) >= 2 |
Qual é o número de telefone e o código postal do endereço 1031 Daugavpils Parkway? | CREATE TABLE address (phone VARCHAR, postal_code VARCHAR, address VARCHAR) | SELECT phone, postal_code FROM address WHERE address = '1031 Daugavpils Parkway' |
Qual cidade tem mais endereços? | CREATE TABLE address (city_id VARCHAR); CREATE TABLE city (city VARCHAR, city_id VARCHAR) | SELECT T2.city, COUNT(*), T1.city_id FROM address AS T1 JOIN city AS T2 ON T1.city_id = T2.city_id GROUP BY T1.city_id ORDER BY COUNT(*) DESC LIMIT 1 |
Quantos endereços há no distrito da Califórnia? | CREATE TABLE address (district VARCHAR) | SELECT COUNT(*) FROM address WHERE district = 'California' |
Qual filme é alugado por 0,99 e tem menos de 3 no inventário? | CREATE TABLE film (title VARCHAR, film_id VARCHAR, rental_rate VARCHAR); CREATE TABLE inventory (film_id VARCHAR); CREATE TABLE film (title VARCHAR, film_id VARCHAR) | SELECT title, film_id FROM film WHERE rental_rate = 0.99 INTERSECT SELECT T1.title, T1.film_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id GROUP BY T1.film_id HAVING COUNT(*) < 3 |
Quantas cidades há na Austrália? | CREATE TABLE country (country_id VARCHAR, country VARCHAR); CREATE TABLE city (country_id VARCHAR) | SELECT COUNT(*) FROM city AS T1 JOIN country AS T2 ON T1.country_id = T2.country_id WHERE T2.country = 'Australia' |
Que países têm pelo menos 3 cidades? | CREATE TABLE country (country VARCHAR, country_id VARCHAR); CREATE TABLE city (country_id VARCHAR) | SELECT T2.country FROM city AS T1 JOIN country AS T2 ON T1.country_id = T2.country_id GROUP BY T2.country_id HAVING COUNT(*) >= 3 |
Encontre todas as datas de pagamento para os pagamentos com um montante superior a 10 e os pagamentos tratados por um funcionário com o nome de Elsa. | CREATE TABLE payment (payment_date VARCHAR, staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, first_name VARCHAR); CREATE TABLE payment (payment_date VARCHAR, amount INTEGER) | SELECT payment_date FROM payment WHERE amount > 10 UNION SELECT T1.payment_date FROM payment AS T1 JOIN staff AS T2 ON T1.staff_id = T2.staff_id WHERE T2.first_name = 'Elsa' |
Quantos clientes têm um valor ativo de 1? | CREATE TABLE customer (active VARCHAR) | SELECT COUNT(*) FROM customer WHERE active = '1' |
Qual é o filme que tem a maior taxa de locação? | CREATE TABLE film (title VARCHAR, rental_rate VARCHAR) | SELECT title, rental_rate FROM film ORDER BY rental_rate DESC LIMIT 1 |
Qual filme tem o maior número de atores ou atrizes? | CREATE TABLE film_actor (film_id VARCHAR); CREATE TABLE film (title VARCHAR, film_id VARCHAR, description VARCHAR) | SELECT T2.title, T2.film_id, T2.description FROM film_actor AS T1 JOIN film AS T2 ON T1.film_id = T2.film_id GROUP BY T2.film_id ORDER BY COUNT(*) DESC LIMIT 1 |
Qual ator de cinema (atriz) estrelou mais filmes? | CREATE TABLE film_actor (actor_id VARCHAR); CREATE TABLE actor (first_name VARCHAR, last_name VARCHAR, actor_id VARCHAR) | SELECT T2.first_name, T2.last_name, T2.actor_id FROM film_actor AS T1 JOIN actor AS T2 ON T1.actor_id = T2.actor_id GROUP BY T2.actor_id ORDER BY COUNT(*) DESC LIMIT 1 |
Quais atores de cinema (atriz) desempenharam um papel em mais de 30 filmes? | CREATE TABLE film_actor (actor_id VARCHAR); CREATE TABLE actor (first_name VARCHAR, last_name VARCHAR, actor_id VARCHAR) | SELECT T2.first_name, T2.last_name FROM film_actor AS T1 JOIN actor AS T2 ON T1.actor_id = T2.actor_id GROUP BY T2.actor_id HAVING COUNT(*) > 30 |
Qual loja possui mais itens? | CREATE TABLE inventory (store_id VARCHAR) | SELECT store_id FROM inventory GROUP BY store_id ORDER BY COUNT(*) DESC LIMIT 1 |
Qual é o montante total de todos os pagamentos? | CREATE TABLE payment (amount INTEGER) | SELECT SUM(amount) FROM payment |
Qual cliente, que fez pelo menos um pagamento, gastou menos dinheiro? | CREATE TABLE payment (customer_id VARCHAR); CREATE TABLE customer (first_name VARCHAR, last_name VARCHAR, customer_id VARCHAR) | SELECT T1.first_name, T1.last_name, T1.customer_id FROM customer AS T1 JOIN payment AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY SUM(amount) LIMIT 1 |
Qual é o nome do gênero do filme "Hunger Roof"? | CREATE TABLE film_category (category_id VARCHAR, film_id VARCHAR); CREATE TABLE film (film_id VARCHAR, title VARCHAR); CREATE TABLE category (name VARCHAR, category_id VARCHAR) | SELECT T1.name FROM category AS T1 JOIN film_category AS T2 ON T1.category_id = T2.category_id JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T3.title = 'HUNGER ROOF' |
Quantos filmes há em cada categoria? | CREATE TABLE film_category (category_id VARCHAR); CREATE TABLE category (name VARCHAR, category_id VARCHAR) | SELECT T2.name, T1.category_id, COUNT(*) FROM film_category AS T1 JOIN category AS T2 ON T1.category_id = T2.category_id GROUP BY T1.category_id |
Qual é o filme que tem mais cópias no inventário? | CREATE TABLE film (title VARCHAR, film_id VARCHAR); CREATE TABLE inventory (film_id VARCHAR) | SELECT T1.title, T1.film_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id GROUP BY T1.film_id ORDER BY COUNT(*) DESC LIMIT 1 |
Qual é o título do filme e o número de inventário do item do inventário que foi alugado com mais frequência? | CREATE TABLE film (title VARCHAR, film_id VARCHAR); CREATE TABLE inventory (inventory_id VARCHAR, film_id VARCHAR); CREATE TABLE rental (inventory_id VARCHAR) | SELECT T1.title, T2.inventory_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id JOIN rental AS T3 ON T2.inventory_id = T3.inventory_id GROUP BY T2.inventory_id ORDER BY COUNT(*) DESC LIMIT 1 |
Quantas línguas estão nestes filmes? | CREATE TABLE film (language_id VARCHAR) | SELECT COUNT(DISTINCT language_id) FROM film |
- Quais são os filmes classificados como R? | CREATE TABLE film (title VARCHAR, rating VARCHAR) | SELECT title FROM film WHERE rating = 'R' |
Onde fica a loja 1? | CREATE TABLE store (address_id VARCHAR); CREATE TABLE address (address VARCHAR, address_id VARCHAR) | SELECT T2.address FROM store AS T1 JOIN address AS T2 ON T1.address_id = T2.address_id WHERE store_id = 1 |
Qual o pessoal que processou o menor número de pagamentos? | CREATE TABLE payment (staff_id VARCHAR); CREATE TABLE staff (first_name VARCHAR, last_name VARCHAR, staff_id VARCHAR) | SELECT T1.first_name, T1.last_name, T1.staff_id FROM staff AS T1 JOIN payment AS T2 ON T1.staff_id = T2.staff_id GROUP BY T1.staff_id ORDER BY COUNT(*) LIMIT 1 |
Em que língua é usado o filme AIRPORT POLLOCK? | CREATE TABLE film (language_id VARCHAR, title VARCHAR); CREATE TABLE LANGUAGE (name VARCHAR, language_id VARCHAR) | SELECT T2.name FROM film AS T1 JOIN LANGUAGE AS T2 ON T1.language_id = T2.language_id WHERE T1.title = 'AIRPORT POLLOCK' |
Quantas lojas há? | CREATE TABLE store (Id VARCHAR) | SELECT COUNT(*) FROM store |
Quantos tipos de classificações diferentes estão listados? | CREATE TABLE film (rating VARCHAR) | SELECT COUNT(DISTINCT rating) FROM film |
Que filmes têm 'Escenas apagadas' como substring no recurso especial? | CREATE TABLE film (title VARCHAR, special_features VARCHAR) | SELECT title FROM film WHERE special_features LIKE '%Deleted Scenes%' |
Quantos itens no inventário tem a loja 1? | CREATE TABLE inventory (store_id VARCHAR) | SELECT COUNT(*) FROM inventory WHERE store_id = 1 |
Quando foi o primeiro pagamento? | CREATE TABLE payment (payment_date VARCHAR) | SELECT payment_date FROM payment ORDER BY payment_date LIMIT 1 |
Onde vive a cliente com o nome de Linda e qual é o e-mail dela? | CREATE TABLE customer (email VARCHAR, address_id VARCHAR, first_name VARCHAR); CREATE TABLE address (address VARCHAR, address_id VARCHAR) | SELECT T2.address, T1.email FROM customer AS T1 JOIN address AS T2 ON T2.address_id = T1.address_id WHERE T1.first_name = 'LINDA' |
Encontre todos os filmes com mais de 100 minutos, ou classificados PG, exceto aqueles que custam mais de 200 para a substituição. | CREATE TABLE film (title VARCHAR, replacement_cost INTEGER, LENGTH VARCHAR, rating VARCHAR) | SELECT title FROM film WHERE LENGTH > 100 OR rating = 'PG' EXCEPT SELECT title FROM film WHERE replacement_cost > 200 |
Qual é o nome e o apelido do cliente que fez o primeiro aluguel? | CREATE TABLE customer (first_name VARCHAR, last_name VARCHAR, customer_id VARCHAR); CREATE TABLE rental (customer_id VARCHAR, rental_date VARCHAR) | SELECT T1.first_name, T1.last_name FROM customer AS T1 JOIN rental AS T2 ON T1.customer_id = T2.customer_id ORDER BY T2.rental_date LIMIT 1 |
Qual é o nome completo do funcionário que alugou um filme a um cliente com o nome de April e o sobrenome Burns? | CREATE TABLE customer (customer_id VARCHAR, first_name VARCHAR, last_name VARCHAR); CREATE TABLE rental (staff_id VARCHAR, customer_id VARCHAR); CREATE TABLE staff (first_name VARCHAR, last_name VARCHAR, staff_id VARCHAR) | SELECT DISTINCT T1.first_name, T1.last_name FROM staff AS T1 JOIN rental AS T2 ON T1.staff_id = T2.staff_id JOIN customer AS T3 ON T2.customer_id = T3.customer_id WHERE T3.first_name = 'APRIL' AND T3.last_name = 'BURNS' |
Qual loja tem mais clientes? | CREATE TABLE customer (store_id VARCHAR) | SELECT store_id FROM customer GROUP BY store_id ORDER BY COUNT(*) DESC LIMIT 1 |
Qual é o montante de pagamento mais elevado? | CREATE TABLE payment (amount VARCHAR) | SELECT amount FROM payment ORDER BY amount DESC LIMIT 1 |
Onde vive a funcionária com o nome de Elsa? | CREATE TABLE staff (address_id VARCHAR, first_name VARCHAR); CREATE TABLE address (address VARCHAR, address_id VARCHAR) | SELECT T2.address FROM staff AS T1 JOIN address AS T2 ON T1.address_id = T2.address_id WHERE T1.first_name = 'Elsa' |
Quais são os primeiros nomes dos clientes que não alugaram nenhum filme após "2005-08-23 02:06:01"? | CREATE TABLE customer (first_name VARCHAR, customer_id VARCHAR, rental_date INTEGER); CREATE TABLE rental (first_name VARCHAR, customer_id VARCHAR, rental_date INTEGER) | SELECT first_name FROM customer WHERE NOT customer_id IN (SELECT customer_id FROM rental WHERE rental_date > '2005-08-23 02:06:01') |
Quantas filiais bancárias há? | CREATE TABLE bank (Id VARCHAR) | SELECT COUNT(*) FROM bank |
Quantos clientes há? | CREATE TABLE bank (no_of_customers INTEGER) | SELECT SUM(no_of_customers) FROM bank |
Encontre o número de clientes nos bancos de Nova Iorque. | CREATE TABLE bank (no_of_customers INTEGER, city VARCHAR) | SELECT SUM(no_of_customers) FROM bank WHERE city = 'New York City' |
Encontre o número médio de clientes em todos os bancos do estado de Utah. | CREATE TABLE bank (no_of_customers INTEGER, state VARCHAR) | SELECT AVG(no_of_customers) FROM bank WHERE state = 'Utah' |
Encontre o número médio de clientes em todos os bancos. | CREATE TABLE bank (no_of_customers INTEGER) | SELECT AVG(no_of_customers) FROM bank |
Encontre a cidade e o estado da filial do banco chamada Morningside. | CREATE TABLE bank (city VARCHAR, state VARCHAR, bname VARCHAR) | SELECT city, state FROM bank WHERE bname = 'morningside' |
Encontre os nomes das agências bancárias no estado de Nova York. | CREATE TABLE bank (bname VARCHAR, state VARCHAR) | SELECT bname FROM bank WHERE state = 'New York' |
Escrever o nome de todos os clientes, ordenados por saldo da conta, em ordem ascendente. | CREATE TABLE customer (cust_name VARCHAR, acc_bal VARCHAR) | SELECT cust_name FROM customer ORDER BY acc_bal |
Escrever o nome de todos os clientes que têm algum empréstimo, ordenado por seu valor total do empréstimo. | CREATE TABLE loan (cust_id VARCHAR, amount INTEGER); CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR) | SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id GROUP BY T1.cust_name ORDER BY SUM(T2.amount) |
Encontre o estado, tipo de conta e pontuação de crédito do cliente cujo número de empréstimo é 0. | CREATE TABLE customer (state VARCHAR, acc_type VARCHAR, credit_score VARCHAR, no_of_loans VARCHAR) | SELECT state, acc_type, credit_score FROM customer WHERE no_of_loans = 0 |
Encontre o número de cidades em que os bancos estão localizados. | CREATE TABLE bank (city VARCHAR) | SELECT COUNT(DISTINCT city) FROM bank |
Encontre o número de estados em que os bancos estão localizados. | CREATE TABLE bank (state VARCHAR) | SELECT COUNT(DISTINCT state) FROM bank |
Quantos tipos diferentes de contas há? | CREATE TABLE customer (acc_type VARCHAR) | SELECT COUNT(DISTINCT acc_type) FROM customer |
Encontre o nome e o saldo da conta do cliente cujo nome inclui a letra a. | CREATE TABLE customer (cust_name VARCHAR, acc_bal VARCHAR) | SELECT cust_name, acc_bal FROM customer WHERE cust_name LIKE '%a%' |
Encontre o saldo total de cada cliente de Utah ou Texas. | CREATE TABLE customer (acc_bal INTEGER, state VARCHAR) | SELECT SUM(acc_bal) FROM customer WHERE state = 'Utah' OR state = 'Texas' |
Encontre o nome dos clientes que têm os dois tipos de contas de poupança e de cheques. | CREATE TABLE customer (cust_name VARCHAR, acc_type VARCHAR) | SELECT cust_name FROM customer WHERE acc_type = 'saving' INTERSECT SELECT cust_name FROM customer WHERE acc_type = 'checking' |
Encontre o nome dos clientes que não têm uma conta poupança. | CREATE TABLE customer (cust_name VARCHAR, acc_type VARCHAR) | SELECT cust_name FROM customer EXCEPT SELECT cust_name FROM customer WHERE acc_type = 'saving' |
Encontre o nome de clientes que não têm um empréstimo com um tipo de hipotecas. | CREATE TABLE loan (cust_id VARCHAR, loan_type VARCHAR); CREATE TABLE customer (cust_name VARCHAR); CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR) | SELECT cust_name FROM customer EXCEPT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE T2.loan_type = 'Mortgages' |
Encontre o nome de clientes que têm empréstimos de ambos os empréstimos hipotecários e de automóveis. | CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR); CREATE TABLE loan (cust_id VARCHAR) | SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE loan_type = 'Mortgages' INTERSECT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE loan_type = 'Auto' |
Encontre o nome dos clientes cuja pontuação de crédito é inferior à média de todas as pontuações de crédito dos clientes. | CREATE TABLE customer (cust_name VARCHAR, credit_score INTEGER) | SELECT cust_name FROM customer WHERE credit_score < (SELECT AVG(credit_score) FROM customer) |
Encontre o nome da filial do banco que tem o maior número de clientes. | CREATE TABLE bank (bname VARCHAR, no_of_customers VARCHAR) | SELECT bname FROM bank ORDER BY no_of_customers DESC LIMIT 1 |
Encontre o nome do cliente que tem a pontuação de crédito mais baixa. | CREATE TABLE customer (cust_name VARCHAR, credit_score VARCHAR) | SELECT cust_name FROM customer ORDER BY credit_score LIMIT 1 |
Encontre o nome, o tipo de conta e o saldo da conta do cliente que tem a pontuação de crédito mais alta. | CREATE TABLE customer (cust_name VARCHAR, acc_type VARCHAR, acc_bal VARCHAR, credit_score VARCHAR) | SELECT cust_name, acc_type, acc_bal FROM customer ORDER BY credit_score DESC LIMIT 1 |
Encontre o nome do cliente que tem o maior montante de empréstimos. | CREATE TABLE loan (cust_id VARCHAR, amount INTEGER); CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR) | SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id GROUP BY T1.cust_name ORDER BY SUM(T2.amount) DESC LIMIT 1 |
Encontre o estado que tem mais clientes. | CREATE TABLE bank (state VARCHAR, no_of_customers INTEGER) | SELECT state FROM bank GROUP BY state ORDER BY SUM(no_of_customers) DESC LIMIT 1 |
Para cada tipo de conta, encontre o saldo médio da conta dos clientes com pontuação de crédito inferior a 50. | CREATE TABLE customer (acc_type VARCHAR, acc_bal INTEGER, credit_score INTEGER) | SELECT AVG(acc_bal), acc_type FROM customer WHERE credit_score < 50 GROUP BY acc_type |
Para cada estado, encontrar o saldo total das contas dos clientes cuja pontuação de crédito seja superior a 100. | CREATE TABLE customer (state VARCHAR, acc_bal INTEGER, credit_score INTEGER) | SELECT SUM(acc_bal), state FROM customer WHERE credit_score > 100 GROUP BY state |
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